Lecture 7
ELE 301: Signals and Systems
Prof. Paul Cuff
Princeton University
Fall 2011-12
Cuff (Lecture 7)
ELE 301: Signals and Systems
Fall 2011-12
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Fall 2011-12
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Introduction to Fourier Transforms
Fourier transform as a limit of the Fourier series
Inverse Fourier transform: The Fourier integral theorem
Example: the rect and sinc functions
Cosine and Sine Transforms
Symmetry properties
Periodic signals and δ functions
Cuff (Lecture 7)
ELE 301: Signals and Systems
Fourier Series
Suppose x(t) is not periodic. We can compute the Fourier series as if x
was periodic with period T by using the values of x(t) on the interval
t ∈ [−T /2, T /2).
ak
=
xT (t) =
Z
1 T /2
x(t)e −j2πkf0 t dt,
T −T /2
∞
X
ak e j2πkf0 t ,
k=−∞
where f0 = 1/T .
The two signals x and xT will match on the interval [−T /2, T /2) but
x̃(t) will be periodic.
What happens if we let T increase?
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Rect Example
For example, assume x(t) = rect(t), and that we are computing the
Fourier series over an interval T ,
f (t) = rect(t)
−1/2
1/2
T
t
The fundamental period for the Fourier series in T , and the fundamental
frequency is f0 = 1/T .
The Fourier series coefficients are
ak =
where sinc(t) =
Cuff (Lecture 7)
1
sinc (kf0 )
T
sin(πt)
πt .
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The Sinc Function
1
-4
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-2
0
2
4
t
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Rect Example Continued
Take a look at the Fourier series coefficients of the rect function (previous
slide). We find them by simply evaluating T1 sinc(f ) at the points f = kf0 .
-8π
1/2
T =2
ω0 = π
1
T =1
ω0 = 2π
-8π
-4π
0
4π
8π
ω = n ω0
-4π
0
T =4
ω0 = π/2
-8π
4π
8π
ω = n ω0
4π
8π
ω = n ω0
1/4
-4π
0
More densely sampled, same sinc() envelope, decreased amplitude.
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Fourier Transforms
Given a continuous time signal x(t), define its Fourier transform as the
function of a real f :
Z ∞
X (f ) =
x(t)e −j2πft dt
−∞
This is similar to the expression for the Fourier series coefficients.
Note: Usually X (f ) is written as X (i2πf ) or X (iω). This corresponds to
the Laplace transform notation which we encountered when discussing
transfer functions H(s).
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ELE 301: Signals and Systems
Fall 2011-12
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We can interpret this as the result of expanding x(t) as a Fourier series in
an interval [−T /2, T /2), and then letting T → ∞.
The Fourier series for x(t) in the interval [−T /2, T /2):
xT (t) =
∞
X
ak e j2πkf0 t
k=−∞
where
ak =
1
T
Z
T /2
x(t)e −j2πkf0 t dt.
−T /2
Define the truncated Fourier transform:
Z T
2
x(t)e −j2πft dt
XT (f ) =
− T2
so that
ak =
Cuff (Lecture 7)
1
1
XT (kf0 ) =
XT
T
T
ELE 301: Signals and Systems
k
T
.
Fall 2011-12
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The Fourier series is then
xT (t) =
∞
X
1
XT (kf0 )e j2πkf0 t
T
k=−∞
The limit of the truncated Fourier transform is
X (f ) = lim XT (f )
T →∞
The Fourier series converges to a Riemann integral:
x(t) =
=
lim xT (t)
T →∞
lim
T →∞
Z
∞
X
k
k
1
XT
e j2π T t
T
T
k=−∞
∞
=
X (f )e j2πft df .
−∞
Cuff (Lecture 7)
ELE 301: Signals and Systems
Fall 2011-12
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Continuous-time Fourier Transform
Which yields the inversion formula for the Fourier transform, the Fourier
integral theorem:
Z
∞
X (f ) =
Z−∞
∞
x(t) =
x(t)e −j2πft dt,
X (f )e j2πft df .
−∞
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ELE 301: Signals and Systems
Fall 2011-12
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Comments:
There are usually technical conditions which must be satisfied for the
integrals to converge – forms of smoothness or Dirichlet conditions.
The intuition is that Fourier transforms can be viewed as a limit of
Fourier series as the period grows to infinity, and the sum becomes an
integral.
R∞
j2πft df is called the inverse Fourier transform of X (f ).
−∞ X (f )e
Notice that it is identical to the Fourier transform except for the sign
in the exponent of the complex exponential.
If the inverse Fourier transform is integrated with respect to ω rather
than f , then a scaling factor of 1/(2π) is needed.
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Cosine and Sine Transforms
Assume x(t) is a possibly complex signal.
Z ∞
X (f ) =
x(t)e −j2πft dt
Z−∞
∞
=
x(t) (cos(2πft) − j sin(2πft)) dt
Z−∞
Z ∞
∞
=
x(t) cos(ωt)dt − j
x(t) sin(ωt) dt.
−∞
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−∞
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Fourier Transform Notation
For convenience, we will write the Fourier transform of a signal x(t) as
F [x(t)] = X (f )
and the inverse Fourier transform of X (f ) as
F −1 [X (f )] = x(t).
Note that
F −1 [F [x(t)]] = x(t)
and at points of continuity of x(t).
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Duality
Notice that the Fourier transform F and the inverse Fourier transform
F −1 are almost the same.
Duality Theorem: If x(t) ⇔ X (f ), then X (t) ⇔ x(−f ).
In other words, F [F [x(t)]] = x(−t).
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Example of Duality
Since rect(t) ⇔ sinc(f ) then
sinc(t) ⇔ rect(−f ) = rect(f )
(Notice that if the function is even then duality is very simple)
1
f (t)
⇔
−2π
2π
t
−1/2 0 1/2
ω
2π
1 f (t)
−2π
F(ω)
⇔
2π
t
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F(ω)
1
−1/2 0 1/2
ELE 301: Signals and Systems
ω
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Generalized Fourier Transforms: δ Functions
A unit impulse δ(t) is not a signal in the usual sense (it is a generalized
function or distribution). However, if we proceed using the sifting property,
we get a result that makes sense:
Z ∞
F [δ(t)] =
δ(t)e −j2πft dt = 1
−∞
so
δ(t) ⇔ 1
This is a generalized Fourier transform. It behaves in most ways like an
ordinary FT.
δ(t)
0
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⇔
t
ELE 301: Signals and Systems
1
0
ω
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Shifted δ
A shifted delta has the Fourier transform
Z ∞
F [δ(t − t0 )] =
δ(t − t0 )e −j2πft dt
−∞
= e −j2πt0 f
so we have the transform pair
δ(t − t0 ) ⇔ e −j2πt0 f
1
δ(t − t0)
0 t0
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e− jωt0
⇔
ω
0
t
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Constant
Next we would like to find the Fourier transform of a constant signal
x(t) = 1. However, direct evaluation doesn’t work:
Z ∞
F [1] =
e −j2πft dt
−∞
∞
e −j2πft =
−j2πf −∞
and this doesn’t converge to any obvious value for a particular f .
We instead use duality to guess that the answer is a δ function, which we
can easily verify.
Z ∞
F −1 [δ(f )] =
δ(f )e j2πft df
−∞
= 1.
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So we have the transform pair
1 ⇔ δ(f )
1
2πδ(ω)
⇔
t
0
ω
0
This also does what we expect – a constant signal in time corresponds to
an impulse a zero frequency.
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Fall 2011-12
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Sinusoidal Signals
If the δ function is shifted in frequency,
Z ∞
F −1 [δ(f − f0 )] =
δ(f − f0 )e j2πft df
−∞
= e j2πf0 t
so
e j2πf0 t ⇔ δ(f − f0 )
1
e jω0t
⇔
0
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t
ELE 301: Signals and Systems
2πδ(ω − ω0)
0 ω0
ω
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Cosine
With Euler’s relations we can find the Fourier transforms of sines and
cosines
1 j2πf0 t
F [cos(2πf0 t)] = F
e
+ e −j2πf0 t
2
h
i
1 h j2πf0 t i
F e
+ F e −j2πf0 t
=
2
1
(δ(f − f0 ) + δ(f + f0 )) .
=
2
so
1
cos(2πf0 t) ⇔ (δ(f − f0 ) + δ(f + f0 )) .
2
1
t
0
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πδ(ω + ω0) πδ(ω − ω0)
cos(ω0t)
⇔
−ω0
0
ω0
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ω
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Sine
Similarly, since sin(f0 t) =
1
j2πf0 t
2j (e
sin(f0 t) ⇔
1
0
− e −j2πf0 t ) we can show that
j
(δ(f + f0 ) − δ(f − f0 )) .
2
sin(ω0t)
t
jπδ(ω + ω0)
⇔
ω0
−ω0
0
ω
− jπδ(ω − ω0)
The Fourier transform of a sine or cosine at a frequency f0 only has energy
exactly at ±f0 , which is what we would expect.
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Fall 2011-12
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